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3 years ago

14

According to Newton's third law, which describes what happens when a 4.5 kg pumpkin is launched from a catapult with a force of

47.5 Newtons?
A.
The pumpkin exerts a force of 43 N back against the catapult.

B.
The pumpkin exerts a force of 52 N back against the catapult.

C.
The pumpkin exerts a force of 4.5 N back against the catapult.

D.
The pumpkin exerts a force of 47.5 N back against the catapult.

1 answer:

max2010maxim [7]3 years ago

6 0

3. Law: Every action has a reaction equal in magnitude and opposite in direction.

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Difference between kilogram and kilometre in points

skelet666 [1.2K]

Answer:

Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.

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3 years ago

If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *

Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

$F_s = \mu_s N$

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

$F_k = \mu_k N$

now we know that

$\mu_k < \mu_s$

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

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3 years ago

All matter in the Universe consists of many substances called elements. <br><br> true or faluse

Mrrafil [7]

The answer to this statement is true!

8 0

3 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

1. Al descomponerse su vehículo una persona tira de su auto con la ayuda de una cuerda con una fuerza de 3500 N que forma un áng

OleMash [197]

Responder:

Fy = 2474,8737

Fx = 2474,8737

Explicación:

Dado que :

Dado:

Fuerza, F = 3500 N

Ángulo formado con la horizontal, θ, = 45 °

Los componentes de una fuerza se pueden descomponer en componentes verticales y horizontales.

El componente vertical Fy; y

El componente horizontal Fx

Fy = Fuerza * sinθ

Fy = 3500 * sin45 °

Fy = 2474,8737

El componente horizontal:

Fx = Fuerza * cosθ

Fy = 3500 * cos45 °

Fy = 2474,8737

8 0

2 years ago